$\begingroup$@Alpha,the OP in this question math.stackexchange.com/questions/313489 don't have problems in figuring out why number of triangles are ${n \choose 3}$+$4{n \choose 4}$+$5{n \choose 5}$+${n \choose 6}$.and the second one you posted is not about combinations.$\endgroup$
– user2838619Jul 27 '14 at 6:45

1 Answer
1

For each selection of any $4$ points on the circumference, you can draw the diagram you have. The line segment that joins the adjacent circumference points, could instead join any of the $4$ pairs of adjacent circumference points, so we have $4$ different triangles for each choice of $4$ circumference points.

There are $\binom{n}{4}$ ways to choose the $4$ points, so Situation $2$ contributes:

$\qquad4 \binom{n}{4}$ triangles.

Situation $3$:

For each selection of any $5$ points on the circumference, you can draw the diagram you have. You could choose any of these $5$ points to be a vertex of a Situation $3$ triangle, so we have $5$ different triangles for each choice of $5$ circumference points.

There are $\binom{n}{5}$ ways to choose the $5$ points, so Situation $3$ contributes:

$\qquad5 \binom{n}{5}$ triangles.

Situation $4$:

For each selection of any $6$ points on the circumference, you can draw the diagram you have. There is only one way to construct that internal triangle given these $6$ circumference points.

There are $\binom{n}{6}$ ways to choose the $6$ points, so Situation $4$ contributes: